\subsection{平方差公式}\label{subsec:6-7}

我们来计算
$$ (a + b) (a - b) \douhao $$
得
\begin{align*}
        & (a + b) (a - b) \\
    ={} & a^2 - ab + ab - b^2 \\
    ={} & a^2 - b^2 \douhao
\end{align*}
由此得到
\begin{center}
    \framebox{\quad $(a + b) (a - b) = a^2 - b^2$。\;}
\end{center}

这就是说， \zhongdian{两个数的和与这两个数的差的积等于这两个数的平方差。} 这个公式就是\zhongdian{平方差公式。}
对于形如两数和与这两数差相乘的乘法， 就可以运用上述公式来计算。 例如， 计算
$$ (1 + 2x) (1 - 2x) \douhao $$
如果把 1 看成 $a$， 把 $2x$ 看成 $b$， 那么
$$ (1 + 2x) (1 - 2x) $$
就是
$$ (a + b) (a - b) $$
的形式。因此，可用平方差公式来计算。即
\begin{align*}
    & (1 + 2x) (1 - 2x) \; = \; 1^2 - (2x)^2 = 1 - 4x^2 \juhao \\[1em]
    \tikz [overlay, >=Stealth] {
        \draw [dashed] (-1em, -1em) rectangle (7em, 1.5em);
        \draw [<->] (.7em, .8em) -- (.7em, 2.5em);
        \draw [<->] (2.4em, .8em) -- (2.4em, 2.5em);
        \draw [<->] (4.1em, .8em) -- (4.1em, 2.5em);
        \draw [<->] (5.9em, .8em) -- (5.9em, 2.5em);
    }
    & (a + b)\phantom{x} (a - b)\phantom{x}  \; = \;
    \tikz [overlay, >=Stealth] {
        \draw [dashed] (-.2em, -1em) rectangle (4em, 1.5em);
        \draw [<->] (.3em, .8em) -- (.3em, 2.5em);
        \draw [<->] (3.2em, .8em) -- (3.2em, 2.5em);
    }
    a^2 - \phantom{()}b^2
\end{align*}\vspace*{1em}

\liti 运用平方差公式计算：
\begin{xiaoxiaotis}

    \xxt{$(3m + 2n) (3m - 2n)$；}

    \xxt{$(b^2 + 2a^3) (2a^3 - b^2)$。}

\resetxxt
\jie \xxt{$\begin{aligned}[t]
        & (3m + 2n) (3m - 2n) \\
    ={} & (3m)^2 - (2n)^2 \\
    ={} & 9m^2 - 4n^2 \fenhao
\end{aligned}$}

\xxt{$\begin{aligned}[t]
    & (b^2 + 2a^3) (2a^3 - b^2) \\
    ={} & (2a^3 + b^2) (2a^3 - b^2) \\
    ={} & (2a^3)^2 - (b^2)^2 \\
    ={} & 4a^6 - b^4 \juhao
\end{aligned}$}

\end{xiaoxiaotis}


\liti 运用平方差公式计算：
\begin{xiaoxiaotis}
\begin{enhancedline}

    \xxt{$\left(-\dfrac{1}{2}x + 2y\right) \left(-\dfrac{1}{2}x - 2y\right)$；}

    \xxt{$(-4a - 1) (4a - 1)$。}

\resetxxt
\jie  \xxt{$\begin{aligned}[t]
        & \left(-\dfrac{1}{2}x + 2y\right) \left(-\dfrac{1}{2}x - 2y\right) \\
    ={} & \left(-\dfrac{1}{2}x\right)^2 - (2y)^2 \\
    ={} & \dfrac{1}{4}x^2 - 4y^2 \fenhao
\end{aligned}$}
\end{enhancedline}

\xxt{$\begin{aligned}[t]
    & (-4a - 1) (4a - 1) \\
    ={} & [(-1) - 4a] [(-1) + 4a] \\
    ={} & (-1)^2 - (4a)^2 \\
    ={} & 1 - 16a^2 \douhao
\end{aligned}$ \\
或 \\
$\begin{aligned}[t]
    & (-4a - 1) (4a - 1) \\
    ={} & - (4a + 1) (4a - 1) \\
    ={} & -[(4a)^2 - 1^2] \\
    ={} & -(16a^2 - 1)\\
    ={} & 1 - 16a^2 \juhao
\end{aligned}$
}

\end{xiaoxiaotis}

\liti 运用平方差公式计算：
\begin{xiaoxiaotis}

    \xxt{$102 \times 98$；}

    \xxt{$(y + 2) (y - 2) (y^2 + 4)$。}

\resetxxt
\jie \xxt{$\begin{aligned}[t]
        & 102 \times 98 \\
    ={} & (100 + 2) (100 - 2) \\
    ={} & 100^2 - 2^2 = 10000 - 4 \\
    ={} & 9996 \fenhao
\end{aligned}$}

\xxt{$\begin{aligned}[t]
        & (y + 2) (y - 2) (y^2 + 4) \\
    ={} & (y^2 - 4) (y^2 + 4) \\
    ={} & (y^2)^2 - 4^2 \\
    ={} & y^4 - 16 \juhao
\end{aligned}$}

\end{xiaoxiaotis}


\lianxi
\begin{xiaotis}

\xiaoti{运用平方差公式计算：}
\begin{xiaoxiaotis}

    \begin{tblr}{columns={18em, colsep=0pt}}
        \xxt{$(x + a) (x - a)$；} & \xxt{$(m - n) (m + n)$；} \\
        \xxt{$(a + 3b) (a - 3b)$；} & \xxt{$(1 - 5y) (1 + 5y)$；} \\
        \xxt{$(2a + 3) (2a - 3)$；} & \xxt{$(-2x^2 + 5) (-2x^2 - 5)$；} \\
        \xxt{$(4x - 5y) (4x + 5y)$；} & \xxt{$\left(\dfrac{2}{3}x - 7y\right) \left(\dfrac{2}{3}x + 7y\right)$。}
    \end{tblr}

\end{xiaoxiaotis}


\xiaoti{运用平方差公式计算：}
\begin{xiaoxiaotis}

    \begin{tblr}{columns={18em, colsep=0pt}}
        \xxt{$103 \times 97$；} & \xxt{$59.8 \times 60.2$；} \\
        \xxt{$(x + 3) (x - 3) (x^2 + 9)$；} & \xxt{$\left(x - \dfrac{1}{2}\right) \left(x^2 + \dfrac{1}{4}\right) \left(x + \dfrac{1}{2}\right)$。}
    \end{tblr}

\end{xiaoxiaotis}

\xiaoti{化简下列各式：}
\begin{xiaoxiaotis}

    \xxt{$(x - y) (x + y) + (2x + y) (2x + y)$；}

    \xxt{$(2a - b) (2a + b) - (3a - 2b) (3a - 2b)$。}

\end{xiaoxiaotis}


\xiaoti{下面各式的计算对不对，为什么？如果不对，应怎样改正？}
\begin{xiaoxiaotis}

    \xxt{$(x - 6) (x + 6) = x^2 - 6$；}

    \xxt{$(2x + 3) (x - 3) = 2x^2 - 9$；}

    \xxt{$(5ab + 1) (5ab - 1) = 25a^2b^2 - 1$。}

\end{xiaoxiaotis}
\end{xiaotis}

